Chap 11.20

11. 20 A nonuniform beam 4. 50 m long and weighing 1 11.20 A nonuniform beam 4.50 m long and weighing 1.00 kN makes an angle of 25° below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it. The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension T in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam

Forces on the Beam Gravity (of Beam) Cable Weight of Lamp Pivot Horizontal Vertical Gravity (weight of) Beam 0 −1kN Cable (tension)  C − C cos65° C sin65° Weight of Lamp 0 − 5kN Pivot Px Py Horizontal Equation: 0 − C cos65°+ 0 + Px = 0 Vertical Equation: − 1kN + C sin65° − 5kN + Py = 0

Torque Torque (fulcrum is Pivot) = F x d = F ∙ d⊥ Gravity (of Beam) + 1kN ∙ 2cos25° Cable − C ∙ 3 Weight of Lamp + 5kN ∙ 4.5 cos25° Pivot distance is 0 0 Equation: + 1kN ∙ 2cos25° − 3C + 5kN ∙ 4.5cos25° + 0 = 0

Solving Torque Eq: + 1kN ∙ 2cos25° − 3C + 5kN ∙ 4.5cos25° + 0 = 0 (2kN+22.5kN) ∙ cos25° = 3C C = 24.5kN cos25° / 3 = 24500 N x .9063/3 = 7401 N Horizontal Eq: 0 − C cos65°+ 0 + Px = 0 Px = C cos65° = 7401 * .4226 = 3128 N Vertical Eq: − 1kN + C sin65° − 5kN + Py = 0 Py = 6kN − C sin65° = 6000N – .9063*7401 N = 6000N – 6708N = – 708 N

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